Is Every Prime Factor of a Composite Number Less Than or Equal to sqrt(n)?

[annoymage]

Homework Statement

If n is a composite number, the n has prime factor nor exceeding sqrt(n)

Homework Equations

n/a

The Attempt at a Solution

what did it mean by "nor exceeding sqrt(n)", is it, it must be less than sqrt(n)?

 

[JonF]

i'd gues it meant "not"

 

[annoymage]

is it suppose to mean

If n is a composite number, then n has prime factor not exceeding sqrt(n) ??

 

[Dick]

is it suppose to mean

If n is a composite number, then n has prime factor not exceeding sqrt(n) ??

Yes. If p is the prime factor show p<=sqrt(n).

 

[annoymage]

can you check my proof please,

since n is composite number, so n>3, so by fundamental theorem of arithmetic n can be written by product of primes, say $$pp_1p_2...p_k=n$$, so let $$p_1p_2...p_k=a\ ,\ where\ a<n$$ and assume $$p \leq a$$ without loss generality, so we get $$pa=n\ ,\ 3<p \leq a<n$$, so suppose $$p>\sqrt{n}$$, then $$a>\sqrt{n}$$ then $$ap>\sqrt{n}\sqrt{n}=n$$ contradiction. is it okay?

but it seems something wrong when i assume $$p \leq a$$, because if $$a \leq p$$ we get $$3<a \leq p<n$$ and i can't conclude $$a>\sqrt{n}$$ right? help T_T

 

[Dick]

can you check my proof please,

since n is composite number, so n>3, so by fundamental theorem of arithmetic n can be written by product of primes, say $$pp_1p_2...p_k=n$$, so let $$p_1p_2...p_k=a\ ,\ where\ a<n$$ and assume $$p \leq a$$ without loss generality, so we get $$pa=n\ ,\ 3<p \leq a<n$$, so suppose $$p>\sqrt{n}$$, then $$a>\sqrt{n}$$ then $$ap>\sqrt{n}\sqrt{n}=n$$ contradiction. is it okay?

but it seems something wrong when i assume $$p \leq a$$, because if $$a \leq p$$ we get $$3<a \leq p<n$$ and i can't conclude $$a>\sqrt{n}$$ right? help T_T

I don't why or how you want to assume p<=a. Or why the condition 3<p. Look, you got that if pa=n the either p<=sqrt(n) or a<=sqrt(n). Split into cases.

 

[annoymage]

if pa=n the either p<=sqrt(n) or a<=sqrt(n). Split into cases.

i don't even tried/think to prove that, but thanks, that's a hint for me ;P, so I start to prove that first

if $$p \leq \sqrt{n}$$ then it is proven, if not, suppose $$a>\sqrt{n}$$ then we have $$ap>\sqrt{n}\sqrt{n}=n$$, contradiction. then

if pa=n the either p<=sqrt(n) or a<=sqrt(n)

.

so If $$p \leq \sqrt{n}$$ then the question is proven,

and If $$a \leq \sqrt{n}$$, so $$p_1p_2...p_k \leq \sqrt{n}$$ then
$$p_1 \leq \sqrt{n}$$, then finished, is it ok now?

 

[Dick]

i don't even tried/think to prove that, but thanks, that's a hint for me ;P, so I start to prove that first

if $$p \leq \sqrt{n}$$ then it is proven, if not, suppose $$a>\sqrt{n}$$ then we have $$ap>\sqrt{n}\sqrt{n}=n$$, contradiction. then .

so If $$p \leq \sqrt{n}$$ then the question is proven,

and If $$a \leq \sqrt{n}$$, so $$p_1p_2...p_k \leq \sqrt{n}$$ then
$$p_1 \leq \sqrt{n}$$, then finished, is it ok now?

Yes, now it looks ok.

 

[annoymage]

k thank you very much ^^